Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Section 1.5 - Equations - 1.5 Exercises - Page 57: 111

Answer

$x=729$ and $x=27$

Work Step by Step

$x^{1/2}-3x^{1/3}=3x^{1/6}-9$ Let $x^{1/6}$ be equal to $u$ $u=x^{1/6}$ $u^{2}=x^{1/3}$ $u^{3}=x^{1/2}$ Rewrite the original equation using $u$: $u^{3}-3u^{2}=3u-9$ Take all terms to the left side: $u^{3}-3u^{2}-3u+9=0$ Factor this equation by grouping terms: $(u^{3}-3u^{2})-(3u-9)=0$ Take out common factor $u^{2}$ from the first parentheses and common factor $3$ from the second one: $u^{2}(u-3)-3(u-3)=0$ Rewrite by factoring out common factor $u-3$: $(u-3)(u^{2}-3)=0$ Set both factors equal to $0$ and solve each individual equation for $u$: $u-3=0$ $u=3$ $u^{2}-3=0$ $u^{2}=3$ $u=\pm\sqrt{3}$ Substitute $u$ back to $x^{1/6}$ and solve for $x$: $u=3$ $x^{1/6}=3$ $x=3^{6}$ $x=729$ $u=\sqrt{3}$ $x^{1/6}=\sqrt{3}$ $x=(\sqrt{3})^{6}$ $x=27$ $u=-\sqrt{3}$ $x^{1/6}=-\sqrt{3}$ $x=(-\sqrt{3})^{6}$ $x=27$ The solutions found are $729$ and $27$. The original equation is true for both of them. The final answer is: $x=729$ and $x=27$
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